What is the solution for y^{''}-y^'-2y=e^x-x ?

The solution for y^{''}-y^'-2y=e^x-x is y=c_{1}e^{2x}+c_{2}e^{-x}-1/4+(-e^x+x)/2

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y^{''}-y^'-2y=e^x-x

y^{^{''}} - y^{^{'}} - 2 y = e^{x} - x

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{1}{4} + \frac{- e ^{x} + x}{2}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = e^{x} - x

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{1}{4} + \frac{- e ^{x} + x}{2}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{1}{4} + \frac{- e ^{x} + x}{2}

y^{^{''}} + y^{^{'}} = e^{t} \cdot cos (t)

y^{^{''}} - (2 + 1)^{2} y = e^{28 t} + sin (2 + 1) t

y^{^{''}} - y^{^{'}} - 2 y = 2 x + 7

y^{^{''''}} + 12 y^{^{''}} + 36 = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 x^{2} e^{5 x}

What is the solution for y^{''}-y^'-2y=e^x-x ?
The solution for y^{''}-y^'-2y=e^x-x is y=c_{1}e^{2x}+c_{2}e^{-x}-1/4+(-e^x+x)/2